如何使用shell脚本获取IP地址?

Beg*_*ner 45 shell bash ubuntu shell-script bash-script

想使用 Shell 脚本获取 IP 地址。在不知道 eth0 或 eth1 或 eth2 的情况下如何获得特定的 IP 地址。

我对获取 localhost 地址不感兴趣,想获取私有 IP 地址

lge*_*get 66

你可以做 :

ip addr | grep 'state UP' -A2 | tail -n1 | awk '{print $2}' | cut -f1 -d'/'

它为您提供第一个私有 IP 地址 ip addr

例如,使用ip addr,我得到:

1: lo: <LOOPBACK,UP,LOWER_UP> mtu 65536 qdisc noqueue state UNKNOWN group default 
    link/loopback 00:00:00:00:00:00 brd 00:00:00:00:00:00
    inet 127.0.0.1/8 scope host lo
       valid_lft forever preferred_lft forever
    inet6 ::1/128 scope host 
       valid_lft forever preferred_lft forever
2: em1: <BROADCAST,MULTICAST,UP,LOWER_UP> mtu 1500 qdisc pfifo_fast state UP group default qlen 1000
    link/ether 00:16:76:de:c1:f1 brd ff:ff:ff:ff:ff:ff
    inet 192.168.0.85/24 brd 192.168.0.255 scope global dynamic em1
       valid_lft 42505sec preferred_lft 42505sec
    inet6 fe80::216:76ff:fede:c1f1/64 scope link 
       valid_lft forever preferred_lft forever
3: virbr0: <NO-CARRIER,BROADCAST,MULTICAST,UP> mtu 1500 qdisc noqueue state DOWN group default 
    link/ether 52:54:00:da:92:d0 brd ff:ff:ff:ff:ff:ff
    inet 192.168.122.1/24 brd 192.168.122.255 scope global virbr0
       valid_lft forever preferred_lft forever
4: virbr0-nic: <BROADCAST,MULTICAST> mtu 1500 qdisc pfifo_fast master virbr0 state DOWN group default qlen 500
    link/ether 52:54:00:da:92:d0 brd ff:ff:ff:ff:ff:ff
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使用之前的命令行,我得到192.168.0.85哪个是em1.

要将其放入 shell 脚本内的变量中,您可以执行var=$(ip addr | grep 'state UP' -A2 | tail -n1 | awk '{print $2}' | cut -f1 -d'/'). 现在,直到脚本结束,$var都会有 IP 地址的值。


cuo*_*glm 40

要列出所有 IP 地址,无论名称如何,请尝试以下操作:

ifconfig | perl -nle 's/dr:(\S+)/print $1/e'
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或者:

ifconfig | awk '/inet addr/{print substr($2,6)}'
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ifconfig如果您只需要特定接口的 IP,请立即指定接口名称(例如 eth0):

ifconfig eth0 | perl -nle 's/dr:(\S+)/print $1/e'
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或者:

ifconfig eth0 | awk '/inet addr/{print substr($2,6)}'
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  • 感谢您指出,看起来我上次的编辑没有保存。有趣的是,我无法编辑自己的答案,所以只需重新发布:`ip route get 1.1.1.1 | awk '{打印 $NF; exit}` # 获取可以访问外部的接口的ip (8认同)

Rob*_*ans 10

来自stackoverflow 的公然复制粘贴,因为我们不能跨站点欺骗。我知道它不是 bash 或 sh,但是此时谁还没有安装 python?

您应该使用netifaces。它旨在在 Mac OS X、Linux 和 Windows 上跨平台。

>>> import netifaces as ni
>>> ni.interfaces()
['lo', 'eth0', 'eth1', 'vboxnet0', 'dummy1']
>>> ni.ifaddresses('eth0')
{17: [{'broadcast': 'ff:ff:ff:ff:ff:ff', 'addr': '00:02:55:7b:b2:f6'}], 2: [{'broadcast': '24.19.161.7', 'netmask': '255.255.255.248', 'addr': '24.19.161.6'}], 10: [{'netmask': 'ffff:ffff:ffff:ffff::', 'addr': 'fe80::202:55ff:fe7b:b2f6%eth0'}]}
>>> 
>>> ni.ifaddresses.__doc__
'Obtain information about the specified network interface.\n\nReturns a dict whose keys are equal to the address family constants,\ne.g. netifaces.AF_INET, and whose values are a list of addresses in\nthat family that are attached to the network interface.'
>>> # for the IPv4 address of eth0
>>> ni.ifaddresses('eth0')[2][0]['addr']
'24.19.161.6'
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用于索引协议的数字来自/usr/include/linux/socket.h(在 Linux 中)...

#define AF_INET         2       /* Internet IP Protocol         */
#define AF_INET6        10      /* IP version 6                 */
#define AF_PACKET       17      /* Packet family                */
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结束复制粘贴

如果你想要的是up和outbound接口的IP,这个有效。

如果您只想枚举接口,还有另一个选项,因为似乎没有人能够准确理解您想要什么:

import netifaces as ni
ints = ni.interfaces()
for i in ints:
  if 'eth' in i:
    try:
      ni.ifaddresses(i)[2][0]['addr']
      print("interface: " + i)
      print("address: " + ni.ifaddresses(i)[2][0]['addr'])
    except:
      pass
  elif 'wlan' in i:
    try:
      ni.ifaddresses(i)[2][0]['addr']
      print("interface: " + i)
      print("address: " + ni.ifaddresses(i)[2][0]['addr'])
    except:
      pass
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